I have three points on a plane A: (-1, 3, 0), B: (3, 2, 4) and C: (1, -1,5) and I determined that the equation for this plane is 11x - 12y - 14z = -47 To get this I figured out AB vector and AC vector and then by cross product got n vector. With this I used point (-1, 3, 0) to get my equation for the plane.
I am now suppose to obtain the area of the triangle determined by these three points. ????
I also need to find the equation for the plane that is through the origin and perpendicular to
the planes x - y + z = 5 and 2x + y - 2z = 7
By using these two planes I got two normal vectors and obtained the cross product of them to get a vector to use with the point (0,0,0) . I got that the plane perpendicular to the ones above containing the origin would have the equation x + 4y + 3z = 0 Could someone confirm this or tell me where I am in error? Thanks to all of you who help on this site. Frostking